When NH3 is treated with oxygen gas, the products obtained are N2(g) and H2O(l). If standard enthalpies of formation at 298 K for NH3(g) and H2O(l) are –46.00 kJ/mol and –286.0 kJ/mol respectively, calculate the enthalpy change of the reaction.

4NH3 + 3O2 ==> 2N2 + 6H2O

dHrxn = (n*dHf products) - (n*dHf reactants)

To calculate the enthalpy change of the reaction, you need to use the concept of Hess's Law and the standard enthalpies of formation.

Hess's Law states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps if the reaction can be expressed as a combination of several steps.

In this case, you can break down the reaction into two steps:
1) The formation of N2(g) from its elements (N2) in their standard states.
2) The formation of H2O(l) from its elements (H2 and O2) in their standard states.

The balanced equation for the reaction is:

4NH3(g) + 5O2(g) -> 4NO2(g) + 6H2O(l)

Using the balanced equation, you can see that the coefficients can be scaled to obtain the formation equations for N2(g) and H2O(l).

1) N2(g) formation equation:
1/2 N2(g) -> N2(g)
The coefficient of N2(g) in the balanced equation is 4, so you need to divide the formation equation by 4.

2) H2O(l) formation equation:
3H2(g) + 1.5O2(g) -> 2H2O(l)
The coefficient of H2O(l) in the balanced equation is 6, so you need to multiply the formation equation by 6.

Next, you need to calculate the enthalpy change using the given standard enthalpies of formation.

1) For N2(g):
The enthalpy change for the formation of N2(g) can be determined using the standard enthalpy of formation:

ΔH1 = [(1/2)N2(g)] - 0 = (1/2) * 0 = 0 kJ/mol

2) For H2O(l):
The enthalpy change for the formation of H2O(l) can be determined using the standard enthalpy of formation:

ΔH2 = [2H2O(l)] - [3H2(g) + 1.5O2(g)]
= [2(-286.0 kJ/mol)] - [3(0) + 1.5(0)]
= -572.0 kJ/mol

Finally, the overall enthalpy change for the reaction can be calculated by summing up the individual enthalpy changes:

ΔH = 4ΔH1 + 6ΔH2
= 4(0 kJ/mol) + 6(-572.0 kJ/mol)
= -3432.0 kJ/mol

Therefore, the enthalpy change of the reaction is -3432.0 kJ/mol.